WBJEE 2019 Maths Paper with Solutions

WBJEE 2019 Mathematics paper solved by the subject experts at BYJU’S is available on this page. The solutions have been provided in a detailed manner and the question paper along with answer key and solutions are in downloadable PDF form for offline viewing at a later stage. Candidates can gain a much clearer understanding of the questions, the paper pattern, weightage of marks as well as the concepts themselves by engaging in solving the question paper. This will help them study productively and prepare effectively for the entrance examination.

WBJEE 2019 - Maths

Question 1: P is the extremity of the latus rectum of ellipse 3x2 + 4y2 = 48 in the first quadrant. The eccentric angle of P is

  1. a. π/8
  2. b. 3π/4
  3. c. π/3
  4. d. 3π/3

Solution:

  1. Answer: c

    Given

    Equation of ellipse is x216+y212=1\frac{x^{2}}{16}+\frac{y^{2}}{12}=1

    Let point P on curve (4cos θ, 23sinθ2\sqrt{3}sin\theta )

    But we know point P is extremity of latus rectum of ellipse

    So, P is (2, 3)

    On comparing point P

    4 cos θ= 2 and 23sinθ2\sqrt{3}sin\theta =3

    Cos θ = (1/2) and sin θ = √3/2 ⇒ θ= π/3


Question 2: The direction ratios of the normal of the plane through the points (1,2,3), (–1,–2,1) and parallel to x22=y+13=z4\frac{x-2}{2}=\frac{y+1}{3}=\frac{z}{4} is

  1. a. (2, 3, 4)
  2. b. (14, –8, –1)
  3. c. (–2, 0, –3)
  4. d. (1, –2, –3)

Solution:

  1. Answer: b

    Normal of the plane passing through the points (1, 2, 3), (–1, –2, 1) and parallel to given line is

    = i^j^k^2341+12+231=i^j^k^234244\begin{vmatrix} \hat{i} &\hat{j} & \hat{k}\\ 2 & 3 & 4\\ 1+1 & 2+2 & -3-1 \end{vmatrix}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 &3 & 4\\ 2 & 4 & -4 \end{vmatrix}

    = 28i^+16j^+2k^-28\hat{i}+16\hat{j}+2\hat{k}

    Direction ratio’s of the normal to plane is < 14, –8, –1>


Question 3: The equation of the plane which bisects the line joining the points (1, 2, 3) and (3, 4, 5) at right angles is

  1. a. x + y + z = 0
  2. b. x + y – z = 9
  3. c. x + y +z = 9
  4. d. x + y – z + 9 = 0

Solution:

  1. Answer: c

    Equation of plane is (3 – 1)x +(4 – 2)y + (5 – 3)z = k …..(1)

    This plane passes through m =[3+12,2+42,5+32]=\left [ \frac{3+1}{2},\frac{2+4}{2},\frac{5+3}{2} \right ]

    = (2, 3, 4)

    ⇒k = 2 × 2 + 2 × 3 + 2 × 4 = 18 (put in eq (1))

    ⇒ Eq. of plane is x + y + z =9


Question 4: The limit of the interior angle of a regular polygon of n sides as n → ∞is

  1. a. π
  2. b. π/8
  3. c. 3π/2
  4. d. 2π/3

Solution:

  1. Answer: a

    We know,

    One interior angle = (n-2) π/n

    limn(n2)πn=π\lim_{n\rightarrow \infty }\frac{(n-2)\pi }{n}=\pi


Question 5: Let f(x) > 0 for all x and f ‘(x)’ exists for all x. If f is the inverse function of h and h(x)=11+logxh'(x)=\frac{1}{1+log x} . Then f ‘(x)’ will be

  1. a. 1 + log (f(x))
  2. b. 1+ f(x)
  3. c. 1 – log (f(x))
  4. d. logf(x)

Solution:

  1. Answer: a

    Given

    ⇒f(x) =h-1(x)

    ⇒h(f(x)) = x

    On differentiating with respect to x

    ⇒h’(f(x)) . f’(x) = 1

    ⇒ f’(x) = [1/ h’(f(x))] = 1 + log (f(x))


Question 6: Consider the function f(x) = cosx2. Then

  1. a. f is of period 2π
  2. b. f is of period √2π
  3. c. f is not periodic
  4. d. f is of periodic π

Solution:

  1. Answer: c

    Given

    f(x) = cosx2

    Let, f(x) is periodic with period T

    We know, f(x + T) = f(x)

    ⇒cos (x + T)2 = cos x2

    ⇒ cos (x + T)2 – cosx2 = 0

    2sin((x+T)2x22)sin((x+T)2+x22)=0-2sin\left ( \frac{(x+T)^{2}-x^{2}}{2} \right )sin\left ( \frac{(x+T)^{2}+x^{2}}{2} \right )=0

    ⇒ (x + T)2 – x2 = nπ or (x + T)2 + x2 = nπ

    Which is not possible because these equations are quadratic equation not identity

    ⇒ f(x) is not periodic


Question 7: limx0+(ex+x)1/x\lim_{x\rightarrow 0^{+}}(e^{x}+x)^{1/x}

  1. a. Does or exist finitely
  2. b. is 1
  3. c. is e2
  4. d. is 2

Solution:

  1. Answer: c

    limx0+(ex+x)1/x(1)\lim_{x\rightarrow 0^{+}}(e^{x}+x)^{1/x}(1^{\infty })

    limex0+(ex+x)1x\lim_{e^{x\rightarrow 0^{+}}}\frac{(e^{x}+x)-1}{x}

    Using binomial expansion of ex =1 + x + x2/2!+…….

    ⇒ e 1+1 =e2


Question 8: Let f(x) be a derivable function, f ‘(x) > f(x) and f(0) = 0. Then

  1. a. f(x) > 0 for all x > 0
  2. b. f(x) < 0 for all x > 0
  3. c. no sign of f (x) can be ascertained
  4. d. f(x) is a constant function

Solution:

  1. Answer: a

    Given

    f ‘(x) > f(x)

    Multiply e–x both sides

    ⇒f’(x). e–x – f(x).e–x> 0

    ⇒ (f(x). e–x)’> 0

    ⇒e –x. f(x) is increasing function

    ex.f(x)>e0.f(0)x>0e^{-x}.f(x)> e^{-0}.f(0)\forall x> 0

    WBJEE 2019 Maths Solutions

    ⇒e –x. f(x) >0 since, f(0) = 0 (given)

    f(x)>0x>0f(x)> 0 \forall x> 0


Question 9: Let f: [1, 3] → R be a continuous function that is differentiable in (1, 3) and f ‘(x) = |f(x)|2 + 4 for all x ∈ (1, 3). Then

  1. a. f(3) – f(1) = 5 is true
  2. b. f(3) – f(1) = 5 is false
  3. c. f(3) – f(1) = 7 is false
  4. d. f(3) – f(1) < 0 only at one point (1, 3)

Solution:

  1. Answer: (b,c)

    f ‘(x) = |f(x)|2 + 4

    Using LMVT theorem

    f(3)f(1)31\frac{f(3)-f(1)}{3-1} = f’ (c) for atleast one c∈ (1, 3)

    ⇒f(3) – f(1) = 2(f(c))2 + 8

    ⇒f(3) – f(1) ≥ 8

    ⇒f(3) – f(1) = 5 (false)

    Similarly f(3) – f(1) = 7 (false)


Question 10: limx0+(xnlnx),n>0\lim_{x\rightarrow 0^{+}}(x^{n}lnx),n> 0

  1. a. does not exist
  2. b. exists and is zero
  3. c. exists and is 1
  4. d. exists and is e–1

Solution:

  1. Answer: b

    limx0+(xnlnx),n>0\lim_{x\rightarrow 0^{+}}(x^{n}lnx),n> 0

    limx0+[lnx1/xn][]\lim_{x\rightarrow 0^{+}}\left [ \frac{lnx}{1/x^{n}} \right ]\left [ \frac{\infty }{\infty } \right ] form

    Using L-Hospital Rule

    limx0+[1/xnxn+1]\lim_{x\rightarrow 0^{+}}\left [ \frac{1/x}{\frac{-n}{x^{n+1}}} \right ]

    limx0+[xnn]=0\lim_{x\rightarrow 0^{+}}\left [ \frac{x^{n}}{-n} \right ]=0


Question 11: If cosxlog(tanx2)dx=sinxlog(tanx2)+f(x)\int \cos xlog\left(\tan \frac{x}{2}\right) d x=\sin x \log \left(\tan \frac{x}{2}\right)+f(x) then f(x) is equal to, (assuming c is a arbitrary real constant)

  1. a. c
  2. b. c – x
  3. c. c + x
  4. d. 2x + c

Solution:

  1. Answer: b

    WBJEE 2019 Maths Paper Solutions


Question 12: y=cos{2tan11x1+x}dxy=\int \cos \left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\} dx is an equation of a family of

  1. a. straight lines
  2. b. circles
  3. c. ellipses
  4. d. parabolas

Solution:

  1. Answer: d

    WBJEE 2019 Maths Paper Solved


Question 13: The value of the integration π/4π/4(λsinx+μsinx1+cosx+γ)dx\int_{-\pi / 4}^{\pi / 4}\left(\lambda|\sin x|+\frac{\mu \sin x}{1+\cos x}+\gamma\right) d x

  1. a. is independent of λ only
  2. b. is independent of μ only
  3. c. in independent of γ only
  4. d. depends on γ, μ and λ

Solution:

  1. Answer: b

    WBJEE 2019 Maths Solved Paper


Question 14: The value of limx01x[yaesin2tdtx+yaesin2tdt]\lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]is equal to

  1. a. esin2ye^{sin^{2}y}
  2. b. e2siny
  3. c. esinye^{\left | siny \right |}
  4. d. ecosec2ye^{cosec^{2}y}

Solution:

  1. Answer: a

    Solved Paper of WBJEE 2019 Maths


Question 15: If 22x.2xdx=A22x+c\int 2^{2^x}.2^{x}dx=A2^{2^x}+c , then A =

  1. a. 1/log x
  2. b. log 2. c. (log2)2
  3. d. 1/(log2)2

Solution:

  1. Answer: d

    22x.2xdx=A22x+c\int 2^{2^x}.2^{x}dx=A2^{2^x}+c---------(1)

    Put 22x = t ⇒(22^x.ln2)(2x.ln2)dx = dt

    dt(ln2)2\Rightarrow \int \frac{dt}{(ln2)^{2}}

    t(ln2)2+C22xln22+C\Rightarrow \frac{t}{(ln2)^{2}}+C\Rightarrow \frac{2^{2x}}{ln2^{2}}+C

    (on comparing with equation (1))

    ⇒A = 1/(ln2)2


Question 16: The value of the integral 11{x2015ex(x2+cosx)+1ex}dx\int_{-1}^{1}\left\{\frac{x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)}+\frac{1}{e^{|x|}}\right\} d x is equal to

  1. a. 0
  2. b. 1 – e–1
  3. c. 2e–1
  4. d. 2(1 – e–1)

Solution:

  1. Answer: d

    Solutions of WBJEE 2019 Maths Paper


Question 17: limx1n{1+nn+3+nn+6+nn+9++nn+3(n1)}\lim _{x \rightarrow \infty} \frac{1}{n}\left\{1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+\sqrt{\frac{n}{n+9}}+\ldots+\sqrt{\frac{n}{n+3(n-1)}}\right\}

  1. a. does not exist
  2. b. is 1
  3. c. is 2
  4. d. is 3

Solution:

  1. Answer: c

    Sample Paper of WBJEE 2019 Maths

    = (2/3) x 3 =2


Question 18: The general solution of the differential equation (1+exy)dx+(1xy)exydy=0\left(1+e^{\frac{x}{y}}\right) d x+\left(1-\frac{x}{y}\right) e^{\frac{x}{y}} d y=0 is (c is an arbitrary constant)

  1. a. x-ye(x/y) = c
  2. b. y-xe(x/y) =c
  3. c. x+ye(x/y)= c
  4. d. y+xe(x/y) = c

Solution:

  1. Answer: c

    Practice Paper of WBJEE 2019 Maths

    Practice Question Paper of WBJEE 2019 Maths


Question 19: General solution of (x+y)2dydx=a2,a0\left ( x+y \right )^{2}\frac{dy}{dx}= a^{2}, a \neq 0 is (c is arbitrary constant)

  1. a. (x/a) = tan(y/a) +c
  2. b. tan xy = c
  3. c. tan (x+y)= c
  4. d. tan(y+c/a) = (x+y/a)

Solution:

  1. Answer: d

    (x+y)2dydx=a2dydx=a2(x+y)2(x+y)^{2}\frac{dy}{dx}=a^{2}\Rightarrow \frac{dy}{dx}=\frac{a^{2}}{(x+y)^{2}}

    Put x + y = t ⇒ 1 + (dy/dx) = (dt/dx) …….(2)

    ⇒ From (1) & (2)

    dx=t2t2+a2dt\Rightarrow dx=\frac{t^{2}}{t^{2}+a^{2}}dt

    dx=[1a2t2+a2]dt\Rightarrow dx=\left [ 1-\frac{a^{2}}{t^{2}+a^{2}} \right ]dt

    Integrating both sides

    ⇒x + c = t – a tan–1 t/a since, t = x + y

    ⇒x = x + y – a tan–1(x+y/a) +c

    atan1(x+ya)=y+cx+ya=tan(y+ca)\Rightarrow atan^{-1}\left ( \frac{x+y}{a} \right )=y+c\Rightarrow \frac{x+y}{a}=tan\left (\frac{y+c}{a} \right )


Question 20. Let P(4, 3) be a point on the hyperbola (x2/a2)-(y2/b2)=1.If the normal at P intersects the x-axis at (16, 0), then the eccentricity of the hyperbola is

  1. a. 52\frac{\sqrt{5}}{2}
  2. b. 2
  3. c. 2\sqrt{2}
  4. d. 3\sqrt{3}

Solution:

  1. Answer: b

    Practice Question Paper with Answers of WBJEE 2019 Maths


Question 21: If the radius of a spherical balloon increases by 0.1% then its volume increases approximately by

  1. a. 0.2%
  2. b. 0.3%
  3. c. 0.4%
  4. d. 0.05%

Solution:

  1. Answer: b

    Volume of spherical balloon V = (4/3)πr3

    ΔVV×100=(43π(r+r10)343πr3)43π3×100=(1+110)31\begin{aligned} \frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=\frac{\left(\frac{4}{3} \pi\left(\mathrm{r}+\frac{\mathrm{r}}{10}\right)^{3}-\frac{4}{3} \pi \mathrm{r}^{3}\right)}{\frac{4}{3} \pi^{3}} \times 100 \\ &=\left(1+\frac{1}{10}\right)^{3}-1 \end{aligned}

    = 1 +(3/10) +(3/100) +(1/1000) –1 ≈ 0.3% approximately


Question 22: The three sides of a right-angled triangle are in G.P. (Geometrical Progression). If the two acute angles be α and β, then tan α and tan β are:

  1. a. 5+12and512\frac{\sqrt{5}+1}{2} and \frac{\sqrt{5}-1}{2}
  2. b. 5+12and512\sqrt{\frac{\sqrt{5}+1}{2}} and \sqrt{\frac{\sqrt{5}-1}{2}}
  3. c. 5and15\sqrt{5}and \frac{1}{\sqrt{5}}
  4. d. 52and25\frac{\sqrt{5}}{2}and \frac{2}{\sqrt{5}}

Solution:

  1. Answer: b

    Let sides are a, ar, ar2 (r > 1)

    Using Pythagoras theorem

    Question Paper with Answers of WBJEE 2019 Maths

    ⇒a2 + a2r2 = a2 r4

    ⇒r4 + r2–1 = 0

    r2=1+52,152\Rightarrow r^{2}=\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2}(not possible)

    r=1+52r=\sqrt{\frac{-1+\sqrt{5}}{2}}

    tanα=1rtan\alpha =\frac{1}{r}

    =251=5+12=\sqrt{\frac{2}{\sqrt{5}-1}}=\sqrt{\frac{\sqrt{5}+1}{2}}

    tanβ=r=512tan\beta =r=\sqrt{\frac{\sqrt{5}-1}{2}}


Question 23: If log26+12x=log2[21/x+8]log_{2}6+\frac{1}{2x}=log_{2}\left [ 2^{1/x}+8 \right ] , then the value of x are

  1. a. (1/4,1/3)
  2. b. (1/4 ,1/2)
  3. c. (-1/4 ,1/2)
  4. d. (1/3 ,-1/2)

Solution:

  1. Answer: b

    log26+12x=log2(21/2+8)log_{2}6+\frac{1}{2x}=log_{2}(2^{1/2}+8)

    log2[21/x+86]=12xlog_{2}\left [ \frac{2^{1/x}+8}{6} \right ]= \frac{1}{2x}

    On taking anti log

    21/x+86=21/2x\Rightarrow \frac{2^{1/x}+8}{6}=2^{1/2x}

    [212x]26.21/2x+8=0\Rightarrow \left [ 2^{\frac{1}{2x}} \right ]^{2}-6.2^{1/2x}+8=0

    ⇒t2-6t+8 =0

    ⇒ t = 4,2

    ⇒ 2(1/2x) = 4,2

    ⇒ 2x = 1, ½

    ⇒ x = ½,1/4


Question 24: Let z be a complex number such that the principal value of argument, arg z > 0. Then arg z – arg(–z) is

  1. a. π/2
  2. b. ± π
  3. c. π
  4. d.– π

Solution:

  1. Answer: c

    ⇒argz – arg(–z)

    ⇒argz – [argz – arg(–1)] since, arg (z1/z2) =argz1 – argz2

    ⇒ argz – [argz – arg(–1)] since, arg(–1) = π

    ⇒ π


Question 25: The general value of the real angle θ, which satisfies the equation

(cos θ + i sin θ)(cos2 θ + i sin2 θ)..........(cos n θ + i sin θ) = 1 is given by (assuming k is an integer)

  1. a. 2kπn+2\frac{2k\pi }{n+2}
  2. b. 4kπn(n+1)\frac{4k\pi }{n(n+1)}
  3. c. 4kπ(n+1)\frac{4k\pi }{(n+1)}
  4. d. 6kπn(n+1)\frac{6k\pi }{n(n+1)}

Solution:

  1. Answer: b

    ⇒ (cos θ + i sin θ) (cos 2θ + i sin 2θ)…………(cos nθ + i sin nθ) = 1

    ⇒ e i θ.e i2 θ………e in θ =1 (using Euler’s formula)

    ⇒ e i θ(1+2+………+n) =1

    eiθn(n+1)2=1e^{i\theta \frac{n(n+1)}{2}}=1

    [n(n+1)θ2]=2kπ\left [ \frac{n(n+1)\theta }{2} \right ]=2k\pi

    θ=4kπn(n+1)\theta =\frac{4k\pi }{n(n+1)}


Question 26: Let a,b,c be real numbers such a + b + c < 0 and the quadratic equation ax2 + bx + c = 0 has imaginary roots. Then

  1. a. a > 0, c > 0
  2. b. a > 0, c < 0
  3. c. a < 0, c > 0
  4. d. a < 0, c < 0

Solution:

  1. Answer: d

    Given

    ⇒ ax2 + bx + c = 0 has imaginary roots ⇒D < 0

    ⇒ a + b + c < 0 ⇒f(1) < 0

    ⇒ f(x) < 0 \forallx∈R and a < 0

    ⇒ f(0) < 0 ⇒ c < 0


Question 27: A candidate is required to answer 6 out of 12 questions which are divided into two parts A and B each containing 6 questions and he/she is not permitted to attempt more than 4 questions from any part. In how many different ways can he/she make up his/her choice of 6 questions?

  1. a. 850
  2. b. 800
  3. c. 750
  4. d. 700

Solution:

  1. Answer: a

    6C2 × 6C4 + 6C3× 6C3 + 6C4 × 6C2

    ⇒ (15)2 + (20)2 + (15)2

    ⇒225 + 400 + 225 = 850


Question 28: There are 7 greetings cards, each of a different colour and 7 envelopes of same 7 colours as that of the cards. The number of ways in which the cards can be put in envelopes, so that exactly 4 of the cards go into envelopes of respective colour is:

  1. a. 7C3
  2. b. 2. 7C3
  3. c. 3! 4C4
  4. d. 3! 7C34C3

Solution:

  1. Answer: b

    7C4 × (De-arrangement of 3 things)

    ⇒ 35 × 2 = 70 = 7C3× 2


Question 29.72n + 16n – 1 (n ∈ N) is divisible by

  1. a. 65
  2. b. 63
  3. c. 61
  4. d. 64

Solution:

  1. Answer: d

    ⇒72n + 16n –1

    ⇒ (8 – 1)2n + 16n– 1

    ⇒ [2nC082n _ 2nC182n-1 + --------2nC 2n-181+2nC2n]+16n-1

    ⇒[64λ\lambda-2n.8+1] + 16n – 1

    ⇒64λ\lambda – 16n + 1 + 1 6n – 1

    ⇒64λ\lambda


Question 30: The number of irrational terms in the expansion of (3 1/8 + 5 1/4)84 is

  1. a. 73
  2. b. 74
  3. c. 75
  4. d. 76

Solution:

  1. Answer: b

    ⇒ (84-n)/8 = rational⇒ n = 4, 12 ……………, 76

    ⇒n/4= rational ⇒ n = 4, 8, 12...……………., 84

    ⇒ n can take total 11 terms

    ⇒Total number of rational terms = 11

    Irrational terms = total terms – rational terms = 85 – 11 = 74


Question 31: Let A be a square matrix of order 3 whose all entries are 1 and let I3 be the identity matrix of order 3. Then the matrix A – 3I3 is

  1. a. invertible
  2. b. orthogonal
  3. c. non-invertible
  4. d. real skew symmetric matrix

Solution:

  1. Answer: c

    A=111111111;I3=100010001A=\begin{vmatrix} 1 & 1 & 1\\ 1& 1 &1 \\ 1 & 1 &1 \end{vmatrix};I_{3}=\begin{vmatrix} 1 & 0 & 0\\ 0&1 &0 \\ 0& 0 &1 \end{vmatrix}

    A3I3=111111111300030003=211121112\Rightarrow A-3I_{3}=\begin{vmatrix} 1 & 1 & 1\\ 1& 1 &1 \\ 1 & 1 &1 \end{vmatrix}-\begin{vmatrix} 3 & 0 & 0\\ 0&3 &0 \\ 0& 0 &3 \end{vmatrix}=\begin{vmatrix} -2 &1 &1 \\ 1& -2 &1 \\ 1& 1 &-2 \end{vmatrix}

    ⇒det. (A –3I3) = –2(3) –1(–3) + 1(3) = 0

    ⇒ A – 3I3 is non invertible matrix


Question 32: If M is any square matrix of order 3 over R and If M’ be the transpose of M, then adj(M’) – adj(M)’ is equal to

  1. a. M
  2. b. M’
  3. c. null matrix
  4. d. identity matrix

Solution:

  1. Answer: c

    We know, for square matrix adj(M’) = (adj M)’

    So, adj (M’) – (adj M)’ = 0 = null matrix


Question 33: If A=[55xx0x5x005]A=\begin{bmatrix} 5 & 5x & x\\ 0 & x & 5x\\ 0 & 0 & 5 \end{bmatrix} and A2=25\left | A^{2} \right |=25,then x\left | x \right | is equal to

  1. a. 1/5
  2. b. 5
  3. c. 52
  4. d. 1

Solution:

  1. Answer: a

    A=55xx0x5x005(given)\Rightarrow A=\begin{vmatrix} 5 & 5x & x\\ 0& x &5x \\ 0 & 0 &5 \end{vmatrix}(given)

    A2=25(given)\left | A^{2} \right |=25(given)

    A=±5(given)\left | A \right |=\pm 5(given)

    A=55xx0x5x005=±5\Rightarrow \left | A \right |=\begin{vmatrix} 5 & 5x & x\\ 0& x &5x \\ 0 & 0 &5 \end{vmatrix}=\pm 5

    25x=±525x=\pm 5

    x=±1/5x=\pm 1/5

    x=1/5\left |x \right |= 1/5


Question 34: Let A and B be two square matrices of order 3 and AB = O3, where O3 denotes the null matrix of order 3. Then

  1. a. must be A = O3, B = O3
  2. b. if A \neqO3, must be B \neq O3
  3. c. if A = O3, must be B \neq O3
  4. d. may be A \neqO3, B \neqO3

Solution:

  1. Answer: d

    Let A=100000000andB=000000001A=\begin{vmatrix} 1 & 0 & 0\\ 0 &0 &0 \\ 0& 0 &0 \end{vmatrix} and B=\begin{vmatrix} 0 &0 &0 \\ 0 & 0 &0 \\ 0 & 0 &1 \end{vmatrix}

    Hence AB = 0 but A \neq0, B \neq0

    So, if AB = 0 then may be A \neq0, B = 0


Question 35: Let P and T be the subsets of X-Y plane defined by

P = {(x,y): x > 0, y > 0 and x2 + y2 = 1}

T = {(x,y): x > 0, y > 0 and x8 + y8< 1}

Then PTP\cap T is

  1. a. the void set ϕ\phi
  2. b. P
  3. c. T
  4. d. P–TC

Solution:

  1. Answer: b

    Let (h, k) satisfies x2 + y2 = 1 then h2 + k2 = 1

    Now h8 + k8 = h8 + (1 – h2)4 = 2h8 –4h6 – 4h2 + 1

    = 2h2 (h2 – 1) (h4 – h2 + 2) + 1

    = –2h2k2 (h4 –h2 + 2) + 1 < 1 \forall h > 0, k > 0

    All solution of x2 + y2 = 1 satisfies x8 + y8< 1

    PTP\cap T =P


Question 36: Let f: R \rightarrowR be defined by f(x) = x2 – (x2/1+x2) for all x ϵ\epsilonR. Then

  1. a. f is one-one but not onto mapping
  2. b. f is onto but not one-one mapping
  3. c. f is both one-one and onto
  4. d. f is neither one-one nor onto

Solution:

  1. Answer: d

    f(x)=x2x21+x2f(x)=x^{2}-\frac{x^{2}}{1+x^{2}}

    f(x)=x41+x2\Rightarrow f(x)=\frac{x^{4}}{1+x^{2}}

    Therefore, f(-x) =f(x)

    Therefore, f(x) is even function, hence it is many one.

    Also f(x) ≥ O \forall x ϵ\epsilonR , hence it is into function

    Therefore, f(x) is neither one-one nor onto


Question 37: Let the relation ρ be defined on R as a ρb if 1 + ab > 0. Then

  1. a. ρ is reflexive only
  2. b. ρ is an equivalence relation
  3. c. ρ is reflexive and transitive but not symmetric
  4. d. ρ is reflexive and symmetric but not transitive

Solution:

  1. Answer: d

    (a, a) ϵ\epsilon because 1 + a2> 0 ⇒ ρ is reflexive

    If 1 + ab > 0 then 1 + ba > 0 ⇒ if (a, b) ϵ\epsilon ρ then (b, a) ϵ\epsilon ρ

    ⇒ ρ is symmetric

    Now (-4, 1/8) ϵ\epsilon ρ and (1/8,5) ϵ\epsilon ρ

    But (–4, 5) ∉ ρ, hence ρ is not transitive


Question 38: A problem in mathematics is given to 4 students whose chances of solving individually are 1/2,1/3 ,1/4 and 1/5. Then probability that the problem will be solved at least by one student is

  1. a. 2/3
  2. b. 3/5
  3. c. 4/5
  4. d. ¾

Solution:

  1. Answer: c

    Probability that no student solves problem is

    = [112][113][114][115]=12×23×34×45=15\left [ 1-\frac{1}{2} \right ]\left [ 1-\frac{1}{3} \right ]\left [ 1-\frac{1}{4} \right ]\left [ 1-\frac{1}{5} \right ]=\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}=\frac{1}{5}

    ⇒ Probability that the problem will be solved by at least one student is = 1 – (1/5) = 4/5


Question 39: If X is a random variable such that σ (X) = 2.6, then σ (1 – 4X) is equal to

  1. a. 7.8
  2. b. –10.4
  3. c. 13
  4. d. 10.4

Solution:

  1. Answer: d

    Given

    σ (x) = 2.6

    We know

    σ (ax + b) = |a| (σ (x))

    So s(–4x + 1) = |–4|( σ (x)) = 4 × 2.6 = 10.4


Question 40: If esinx – e–sinx – 4 = 0, then the number of real values of x is

  1. a. 0
  2. b. 1
  3. c. 2
  4. d. 3

Solution:

  1. Answer: a

    esinx – e–sinx = 4

    Let esinx = t

    ⇒ t2 –4 t –1 = 0

    t=2+5,25t=2+\sqrt{5},2-\sqrt{5}

    We know that t is real positive number

    t=2+5t=2+\sqrt{5}

    esinx=2+5e^{sinx}=2+\sqrt{5}

    ⇒ [e-1,e1] ∉ 2+52+\sqrt{5}

    ⇒ hence no solution exist


Question 41: The angles of a triangle are in the ratio 2 : 3 : 7 and the radius of the circumscribed circle is 10 cm. The length of the smallest side is

  1. a. 2 cm
  2. b. 5 cm
  3. c. 7cm
  4. d. 10 cm

Solution:

  1. Answer: d

    Let, angles are 2x, 3x, 7x

    We know

    ⇒2x + 3x + 7x = 180° ⇒ 12x = 180° ⇒ x = 15°

    ⇒ angles are – 30°, 45°, 75°

    ⇒ length of smallest side a

    ⇒a/sin A=2R

    ⇒a = 2Rsin A

    ⇒a = 2 × 10 × sin 30° = 10


Question 42: A variable line passes through a fixed point (x1, y1) and meets the axes at A and B. If the rectangle OAPB be completed, the locus of P is, (O being the origin of the system of axes)

  1. a. (y – y1)2 = 4(x – x1)
  2. b. (x1/x) + (y1/y) = 1
  3. c. x2 + y2 = x12 + y12
  4. d. x22x12+y22y12=1\frac{x^{2}}{2x_{1}^{2}}+\frac{y^{2}}{2y_{1}^{2}}=1

Solution:

  1. Answer: b

    Let P(h, k) then A is (h, 0) & B is (0, k)

    Equation of AB is xh+yk=1\frac{x}{h}+\frac{y}{k}=1which passes through (x1, y1) is

    x1h+y1k=1\frac{x_{1}}{h}+\frac{y_{1}}{k}=1

    ⇒ Required locus is x1x+y1y=1\frac{x_{1}}{x}+\frac{y_{1}}{y}=1


Question 43: A straight line through the point (3, -2) is inclined at an angle 60° to the line x + y = 1. If it intersects the X-axis, then its equation will be

  1. a. y+x3+2+33=0y+x\sqrt{3}+2+3\sqrt{3}=0
  2. b. yx3+2+33=0y-x\sqrt{3}+2+3\sqrt{3}=0
  3. c. yx3233=0y-x\sqrt{3}-2-3\sqrt{3}=0
  4. d. xx3+233=0x-x\sqrt{3}+2-3\sqrt{3}=0

Solution:

  1. Answer: b

    L1: 3x+y=1m1=3\sqrt{3}x+y=1\Rightarrow m_{1}=\sqrt{3}

    θ = 60° (given)

    tanθ=m1m21+m1m2\Rightarrow tan\theta =\left | \frac{m_{1}-m_{2}}{1+m_{1}m_{2}} \right |

    ±3=3m213m2\pm \sqrt{3}=\frac{-\sqrt{3}-m_{2}}{1-\sqrt{3}m_{2}}

    ⇒ m2 =√3,0 (not possible because lines are not parallel)

    ⇒m2 =√3

    So, equation of line passing through (3, –2) and the slope √3 is

    ⇒ y + 2 =√3 (x – 3)

    ⇒y-√3x+2+3√3=0


Question 44: A variable line passes through the fixed point (α, β). The locus of the foot of the perpendicular from the origin on the line is

  1. a. x2 + y2 – αx – βy = 0
  2. b. x2 – y2 + 2αx + 2βy = 0
  3. c. ax+bx±(α2+β2)=0ax+bx\pm \sqrt{(\alpha ^{2}+\beta ^{2})}=0
  4. d. x2α2+y2β2=1\frac{x^{2}}{\alpha ^{2}}+\frac{y^{2}}{\beta ^{2}}=1

Solution:

  1. Answer: a

    MPQ • MOP = –1

    WBJEE 2019 Maths Paper with Answers

    (kβhα)(kh)=1\left ( \frac{k-\beta }{h-\alpha } \right )\left ( \frac{k}{h} \right )=-1

    ⇒k2 – kβ = – (h2 –hα)

    ⇒ h2 + k2 –hα– kβ = 0

    ⇒Required locus is x2 + y2 –α x –β y = 0


Question 45: If the point of intersection of the lines 2ax + 4ay + c = 0 and 7bx + 3by – d = 0 lies in the 4th quadrant and is equidistant from the two axes, where a, b, c and d are non-zero numbers, then ad: bc equals to

  1. a. 2 : 3
  2. b. 2 : 1
  3. c. 1 : 1
  4. d. 3 : 2

Solution:

  1. Answer: b

    Let point on 4th quadrant which is equidistant from both the axis is (α, –α)

    ⇒ L1: 2aα – 4a α + c = 0 ⇒α=c/2a …………(1)

    ⇒L2: 7bα – 3bα – d = 0⇒ α = d/4b …………..(2)

    From equation (1) & (2)

    ⇒ (c/2a) = (d/4b)

    ⇒ (4/2) = ad/cb

    ⇒ ad : bc = 2 : 1


Question 46: A variable circle passes through the fixed point A(p, q) and touches x-axis. The locus of the other end of the diameter through A is

  1. a. (x – p)2 = 4qy
  2. b. (x – q)2 = 4py
  3. c. (y – p)2 = 4qx
  4. d. (y – q)2 = 4px

Solution:

  1. Answer: a

    Let other end is (h, k) then centre equal to (p+h2,q+k2)\left ( \frac{p+h}{2},\frac{q+k}{2} \right )

    Because circle touches x-axis hence radius = q+k2\left | \frac{q+k}{2} \right |

    We know

    ⇒ (AB) Diameter = 2r

    (hP)2+(kq)2=2q+k2\sqrt{(h-P)^{2}+(k-q)^{2}}=2\left | \frac{q+k}{2} \right |

    WBJEE 2019 Maths Paper with Solutions

    On squaring both sides

    ⇒ (h – p)2 + (k – q)2 = (q + k)2

    ⇒ (h – p)2 = (k + q)2 – (k – q)2

    ⇒ Required locus is = (x – p)2 = 4qy


Question 47: If P(0, 0), Q(1, 0) and R [12,32]\left [ \frac{1}{2},\frac{\sqrt{3}}{2} \right ] are three given points, then the centre of the circle for which the lines PQ, QR and RP are the tangents is

  1. a. [12,14]\left [ \frac{1}{2},\frac{1}{4} \right ]
  2. b. [12,34]\left [ \frac{1}{2},\frac{\sqrt{3}}{4} \right ]
  3. c. [12,123]\left [ \frac{1}{2},\frac{1}{2\sqrt{3}} \right ]
  4. d. [12,13]\left [ \frac{1}{2},\frac{-1}{\sqrt{3}} \right ]

Solution:

  1. Answer: c

    ΔPQR is equilateral triangle

    So, incentre is same as centroid

    ⇒ incentre = centroid = centre of circle =[1+12+03,0+0+323]=[12,123,]\left [ \frac{1+\frac{1}{2}+0}{3},\frac{0+0+\frac{\sqrt{3}}{2}}{3} \right ]=\left [ \frac{1}{2},\frac{1}{2\sqrt{3}}, \right ]


Question 48: For the hyperbola x2cos2αy2sin2α=1\frac{x^{2}}{cos^{2}\alpha }-\frac{y^{2}}{sin^{2}\alpha }=1, which of the following remains fixed when α varies?

  1. a. Directrix
  2. b. Vertices
  3. c. foci
  4. d. Eccentricity

Solution:

  1. Answer: c

    Focus are = (±a2+b2,0)=(±cos2α+sin2α,0)(\pm \sqrt{a^{2}+b^{2},0})=(\pm \sqrt{cos^{2}\alpha +sin^{2}\alpha },0)

    ⇒ focus are (±1,0), which is independent of α

    ⇒focus are fixed


Question 49: S and T are the foci of an ellipse and B is the end point of the minor axis. If STB is equilateral triangle, the eccentricity of the ellipse is

  1. a. 1/4
  2. b. 1/3
  3. c. 1/2
  4. d. 2/3

Solution:

  1. Answer: c

    S = (ae, 0)

    T = (–ae, 0)

    B =(0,b)

    ⇒ (ST)2 = (SB)2

    ⇒ (2ae)2 = (ae)2 + b2

    ⇒3(ae)2 = b2

    ⇒b2 = 3(a2 – b2)

    b2a2=34\frac{b^{2}}{a^{2}}=\frac{3}{4}

    e=1b2a2e=\sqrt{1-\frac{b^{2}}{a^{2}}}

    e=134e=\sqrt{1-\frac{3}{4}}

    e=1/2


Question 50: The equation of the directrices of the hyperbola 3x2 – 3y2 – 18x + 12y + 2 = 0 is

  1. a. x=3±136x=3\pm \sqrt{\frac{13}{6}}
  2. b. x=3±613x=3\pm \sqrt{\frac{6}{13}}
  3. c. x=6±133x=6\pm \sqrt{\frac{13}{3}}
  4. d. x=6±313x=6\pm \sqrt{\frac{3}{13}}

Solution:

  1. Answer: a

    Equation of hyperbola is 3(x2 – 6x) – 3 (y2 –4y) + 2 = 0

    ⇒ (x – 3)2 – (y – 2)2 = 13/3

    (x3)2133(y2)2133=1e=2\Rightarrow \frac{(x-3)^{2}}{\sqrt{\frac{13}{3}}}-\frac{(y-2)^{2}}{\sqrt{\frac{13}{3}}}=1\Rightarrow e=\sqrt{2}

    We know equation of directrix is X = ±ae\pm \frac{a}{e}

    x3=±13/32x-3=\pm \frac{\sqrt{13/3}}{\sqrt{2}}

    x=3±13/32x=3\pm \frac{\sqrt{13/3}}{\sqrt{2}}


Question 51: The graphs of the polynomial x2 – 1 and cos x intersect

  1. a. at exactly two points
  2. b. at exactly 3 points
  3. c. at least 4 but at finitely many points
  4. d. at infinitely many points

Solution:

  1. Answer: a

    y = x2 – 1 & y = cos x

    intersect at exactly two points

    WBJEE 2019 Maths Sample Paper


Question 52: A point is in motion along a hyperbola y = 10/x so that its abscissa x increases uniformly at a rate of 1 unit per second. Then, the rate of change of its ordinate, when the point passes through (5, 2)

  1. a. increases at the rate of 1/2 unit per second
  2. b. decreases at the rate of 1/2 unit per second
  3. c. decreases at the rate of 2/5 unit per second
  4. d. increases at the rate of 2/5 unit per second

Solution:

  1. Answer: c

    Given

    (dx/dt) = 1 unit per second

    y = 10/x ………(1)

    differentiating with respect to x

    WBJEE 2019 Maths Sample Question Paper

    ⇒Ordinate decreases at rate unit per second.


Question 53: Let a = min{x2 + 2x + 3: x ϵ\epsilon R} and b = limx01cosθθ2\lim_{x\rightarrow 0}\frac{1-cos\theta }{\theta ^{2}} . Then r=0narbnr\sum_{r=0}^{n}a^{r}b^{n-r} is

  1. a. 2n+113.2n\frac{2^{n+1}-1}{3.2^{n}}
  2. b. 2n+1+13.2n\frac{2^{n+1}+1}{3.2^{n}}
  3. c. 4n+113.2n\frac{4^{n+1}-1}{3.2^{n}}
  4. d. (1/2)(2n – 1)

Solution:

  1. Answer: c

    a = min [x2 + 2x + 3; x ϵ\epsilon R]

    a = min [(x+1)2+2]

    ⇒ a = 2

    b=limθ01cosθθ2=122.2sin2θ2θ222=12b=\lim_{\theta \rightarrow 0}\frac{1-cos\theta }{\theta ^{2}}=\frac{1}{2^{2}}.\frac{2sin^{2}\frac{\theta }{2}}{\frac{\theta ^{2}}{2^{2}}}=\frac{1}{2}

    ⇒b=1/2

    Now,

    r=0narbnr=a0bn+abn1+a2bn2+.........+anb0\sum_{r=0}^{n}a^{r}b^{n-r}=a^{0}b^{n}+ab^{n-1}+a^{2}b^{n-2}+.........+a^{n}b^{0}

    =(12)n+22n1+222n2+.........+2n=\left ( \frac{1}{2} \right )^{n}+\frac{2}{2^{n-1}}+\frac{2^{2}}{2^{n-2}}+.........+2^{n}

    =(12)n[1+4+42+.......+4n]=\left ( \frac{1}{2} \right )^{n}\left [ 1+4+4^{2}+.......+4^{n} \right ]

    12n[4n+1141]\frac{1}{2^{n}}\left [ \frac{4^{n+1}-1}{4-1} \right ]

    [4n+113.2n]\left [ \frac{4^{n+1}-1}{3.2^{n}} \right ]


Question 54: Let a > b >0 and I(n) = a1/n – b1/n, J(n) = ((a – b)1/n for all n ≥ 2. then

  1. a. I(n) < J(n)
  2. b. I(n) > J(n)
  3. c. I(n) = J(n)
  4. d. I(n) + J(n) = 0

Solution:

  1. Answer: a

    Given

    ⇒a > b > 0

    ⇒ a > b

    ⇒ (b/a) <1

    Question Paper of WBJEE 2019 Maths

    = I(n) < J(n)


Question 55: Let α^,β^,γ^\hat{\alpha },\hat{\beta },\hat{\gamma } be three unit vectors such thatα^×(β^×γ^)=12(β^×γ^)\hat{\alpha }\times (\hat{\beta }\times \hat{\gamma })=\frac{1}{2}(\hat{\beta }\times \hat{\gamma }) where α^×(β^×γ^)=(α^.γ^)β^(α^.β^)γ^\hat{\alpha }\times (\hat{\beta }\times \hat{\gamma })=(\hat{\alpha }.\hat{\gamma })\hat{\beta }-(\hat{\alpha }.\hat{\beta })\hat{\gamma } .If β^\hat{\beta }is not parallel to , then the angle between α and β is

  1. a. 5π/6
  2. b. π/6
  3. c. π/3
  4. d. 2π/3

Solution:

  1. Answer: d

    α^=β^=γ^=1(given)\left |\hat{\alpha } \right |= \left | \hat{\beta } \right |=\left | \hat{\gamma }\right |=1 (given)

    Now,

    α^×(β^×γ^)=12(β^×γ^)\hat{\alpha }\times (\hat{\beta }\times \hat{\gamma })=\frac{1}{2}(\hat{\beta }\times \hat{\gamma })

    On comparing both sides

    α^.β^=12\Rightarrow -\hat{\alpha }.\hat{\beta }=\frac{1}{2}

    α^.β^=12\Rightarrow \hat{\alpha }.\hat{\beta }=\frac{-1}{2}

    α^.β^cosθ=12\left | \hat{\alpha } \right | .\left | \hat{\beta } \right |cos\theta =\frac{-1}{2}

    θ=2π3\theta =\frac{2\pi }{3}


Question 56: The position vectors of the points A, B, C and D are 3i^2j^k^3\hat{i}-2\hat{j}-\hat{k}, 2i^3j^+2k^2\hat{i}-3\hat{j}+2\hat{k}, 5i^j^+2k^5\hat{i}-\hat{j}+2\hat{k} and 4i^j^+λk^4\hat{i}-\hat{j}+\lambda \hat{k}respectively. If the points A, B, C and D lie on a plane, the value of λ is

  1. a. 0
  2. b. 1
  3. c. 2
  4. d. – 4

Solution:

  1. Answer: d

    Given

    A, B, C, & D are on a plane

    Since, AB,AC,AD\vec{AB}, \vec{AC},\vec{AD}are coplanar

    Since, [AB,AC,AD]=0[\vec{AB},\vec{AC},\vec{AD}]=0

    113213111λ=0\Rightarrow \begin{vmatrix} 1 & 1 & -3\\ -2& -1 &-3 \\ -1& -1 & -1-\lambda \end{vmatrix}=0

    R3R3+R1R_{3}\rightarrow R_{3}+R_{1}

    113213004λ=0\Rightarrow \begin{vmatrix} 1 & 1 & -3\\ -2& -1 &-3 \\ 0& 0 & -4-\lambda \end{vmatrix}=0

    ⇒ (-4-λ) (-1 + 2) = 0

    ⇒λ = -4


Question 57: A particle starts at the origin and moves 1 unit horizontally to the right and reaches P1, then it moves ½ unit vertically up and reaches P2, then it moves 1/4 unit horizontally to right and reaches P3, then it moves 1/8 unit vertically down and reaches P4, then it moves 1/16 unit horizontally to right and reaches P5 and so on. Let Pn = (xn, yn) and xn =α and yn = β. Then (α,β) is

  1. a. (2, 3)
  2. b. [43,25]\left [ \frac{4}{3},\frac{2}{5} \right ]
  3. c. [25,1]\left [ \frac{2}{5},1 \right ]
  4. d. [43,3]\left [ \frac{4}{3},3 \right ]

Solution:

  1. Answer: b

    Question Paper of WBJEE 2019 Mathematics


Question 58: For any non-zero complex number z, the minimum value of |z| + |z – 1| is

  1. a. 1
  2. b. 1/2
  3. c. 0
  4. d. 3/2

Solution:

  1. Answer: a

    Using inequality |A| + |B|≥ |A – B |

    ⇒|z| +|z – 1|≥ |z –(z – 1)|

    ⇒ |z| + |z–1| ≥ 1

    ⇒ minimum value of |z| + |z – 1| is 1


Question 59: The system of equations

λx + y + 3z = 0

2x + µy – z = 0

5x + 7y + z = 0

Has infinitely many solutions in R. Then,

  1. a. λ = 2, µ = 3
  2. b. λ = 1, µ = 2
  3. c. λ = 1, µ = 3
  4. d. λ = 3, µ = 1

Solution:

  1. Answer: c

    For infinitely many solution Δ= 0

    λ132μ1571=0\Rightarrow\begin{vmatrix} \lambda &1 &3 \\ 2 & \mu & -1\\ 5& 7 &1 \end{vmatrix}=0

    λ μ + 7 λ –7 + 42 – 15 μ =0

    ⇒ (λ – 15) (μ + 7)+ 140 =0

    Now check from options

    ⇒ (λ, μ) = (1, 3)


Question 60: Let f: X \rightarrowY and A, B are non-void subsets of Y, then (where the symbols have their usual interpretation)

  1. a. f–1(A) – f–1(B) \supset f–1(A – B) but the opposite does not hold
  2. b. f–1(A) – f–1(B) \subset f–1(A – B) but the opposite does not hold
  3. c. f–1(A – B) = f–1(A) – f–1(B)
  4. d. f–1(A – B) = f–1(A) \cup f–1(B)

Solution:

  1. Answer: c

    We can see that pre-image of A – B i.e. f–1(A – B) will be f–1(A) – f–1(B)

    Therefore, f–1 (A – B) = f–1(A) – f–1(B)

    Question with Solutions of WBJEE 2019 Mathematics Paper


Question 61: Let S, T, U be three non-void sets and f : S \rightarrowT, g : T \rightarrowU be so that gof : S \rightarrowU is surjective. Then

  1. a. g and f are both surjective
  2. b. g is surjective, f may not be so
  3. c. f is surjective, g may not be so
  4. d. f and g both may not be surjective

Solution:

  1. Answer: b

    Obvious g is surjective otherwise gof cannot be surjective but there is no need of “f” to be surjective.

    Example.

    Solved Question of WBJEE 2019 Mathematics Paper

    Hence f(x) is not surjective still gof is surjective


Question 62: The polar coordinate of a point P is [2,π4]\left [ 2,\frac{-\pi }{4} \right ] . The polar coordinate of the point Q, which is such that the line joining PQ is bisected perpendicularly by the initial line, is

  1. a. [2,π4]\left [ 2,\frac{\pi }{4} \right ]
  2. b. [2,π6]\left [ 2,\frac{\pi }{6} \right ]
  3. c. [2,π4]\left [ -2,\frac{\pi }{4} \right ]
  4. d. [2,π6]\left [ -2,\frac{\pi }{6} \right ]

Solution:

  1. Answer: a

    line joining PQ is bisected by x –axis

    So point Q is [2,π4]\left [ 2,\frac{\pi }{4} \right ]

    Solved Question Paaper of WBJEE 2019 Mathematics


Question 63: The length of conjugate axis of a hyperbola is greater than the length of transverse axis. Then the eccentricity e is

  1. a. = √2
  2. b. > √2
  3. c. < √2
  4. d.1/√2

Solution:

  1. Answer: b

    b > a (given)

    On squaring both sides

    b2a2>1\Rightarrow \frac{b^{2}}{a^{2}}> 1

    On adding 1 in both sides

    1+b2a2>21+\frac{b^{2}}{a^{2}}> 2

    Taking root both sides

    1+b2a2>2\Rightarrow \sqrt{1+\frac{b^{2}}{a^{2}}}> \sqrt{2}

    e>2\Rightarrow e> \sqrt{2}


Question 64: The value of is limx0xp[qx]\lim_{x\rightarrow 0}\frac{x}{p}\left [ \frac{q}{x} \right ]

  1. a. [q]p\frac{\left [ q \right ]}{p}
  2. b. 0
  3. c. 1
  4. d. ∞

Solution:

  1. Answer: a

    WBJEE 2019 Mathematics Solved Question Paper


Question 65: Let f(x) = x4 – 4x3 + 4x2 + c, cϵ\epsilon R. Then

  1. a. f(x) has infinitely many zeroes in (1, 2) for all c
  2. b. f(x) has exactly one zero in (1, 2) if –1 < c < 0
  3. c. f(x) has double zeroes in (1, 2) if –1 < c < 0
  4. d. whatever be the value of c, f(x) has no zero in (1, 2)

Solution:

  1. Answer: b

    f(x) = x4 – 4x3 + 4x2 + c,

    Using IVT theorem

    For atleast one root/zero = f(1) . f(2) < 0

    = (1 + c) c< 0

    = c ϵ\epsilon (–1, 0)

    WBJEE 2019 Mathematics Question Paper Solved

    f(x) = x2(x – 2)2 + c

    ⇒f(x) has exactly one zero in (1, 2) if (–1, 0)


Question 66: Let f and g be differentiable on the interval I and let a, b ϵ\epsilon I, a < b. Then

  1. a. If f(a) = 0 = f(b), the equation f'(x) + f(x)g'(x) = 0 is solvable in (a, b)
  2. b. If f(a) = 0 = f(b), the equation f'(x) + f(x)g'(x) = 0 may not be solvable in (a, b)
  3. c. If g(a) = 0 = g(b), the equation g'(x) + kg(x) = 0 is solvable in (a, b), k ϵ\epsilonR
  4. d. If g(a) = 0 = g(b), the equation g'(x) + kg(x) = 0 may not be solvable in (a, b), k ϵ\epsilon R

Solution:

  1. Answer: (a,c)

    f(a) = 0 = f(b) ⇒f’(a)f’ (b) < 0

    Let h(x) = f’ (x) + f(x) g’ (x) …………..(1)

    Put x = a in equation (1)

    h(a) = f’(a) since f(a) = 0

    Put x = b in equation (1)

    h(b) = f’ (b) since f(b) = 0

    Therefore, h(a) . h(b) < 0 ⇒ h(x) = 0 has roots between (a, b)

    Similarly, g’ (x) + kg(x) = 0 has roots between (a, b) as g(a) = 0 = g(b)


Question 67: Consider the function f(x) = (x3/4) – sin πx + 3

  1. a. f(x) does not attain value within the interval [–2, 2]
  2. b. f(x) takes on the value 2 in the interval [–2, 2]
  3. c. f(x) takes on the value 3 in the interval [–2, 2]
  4. d. f(x) takes no value p, 1 < p < 5 in the interval [–2, 2]

Solution:

  1. Answer:(b,c)

    f(x)=x34sinπx+3f(x)=\frac{x^{3}}{4}-sin\pi x+3

    f(–2) = 1

    f(2) = 5

    since function is continuous

    By intermediate value theorem, f(x) takes all values between 1 to 5


Question 68:Let In=01xntan1xdxI_{n}=\int_{0}^{1}x^{n}tan^{-1}xdx . If anIn+2 + bnIn = cn for all n ≥ 1, then

  1. a. a1, a2, a3 are in G.P.
  2. b. b1, b2, b3 are in A.P.
  3. c. c1, c2, c3 are in H.P.
  4. d. a1, a2, a3 are in A.P

Solution:

  1. Answer: (b,d)

    In=01xntan1xdxI_{n}=\int_{0}^{1}x^{n}tan^{-1}xdx

    Using Integration by parts

    In=tan1xxn+1n+10101xn+1n+1(11+x2)dx\Rightarrow {I}_{n}=\left.\tan ^{-1} {x} \frac{{x}^{{n}+1}}{n+1}\right|_{0} ^{1}-\int_{0}^{1} \frac{{x}^{n+1}}{n+1} \cdot\left(\frac{1}{1+{x}^{2}}\right) dx

    (n+1)In=π401xn+11+x2dx\Rightarrow(n+1) I_{n}=\frac{\pi}{4}-\int_{0}^{1} \frac{x^{n+1}}{1+x^{2}} d x----------(1)

    Put n → n + 2 in equation (1)

    (n+3)In+2=π401xn+31+x2dx\Rightarrow(n+3) I_{n+2}=\frac{\pi}{4}-\int_{0}^{1} \frac{x^{n+3}}{1+x^{2}} d x--------(2)

    From equation(1) + (2)

    ⇒ (n +1)In + (n + 3) In+2 = π21n+2\frac{\pi }{2}-\frac{1}{n+2}

    ⇒ on comparing with given equation

    ⇒an = n + 1, bn = n+ 3, cn=π21n+2c_{n}=\frac{\pi }{2}-\frac{1}{n+2}


Question 69: Two particles A and B move from rest along a straight line with constant accelerations f and h respectively. If A takes m seconds more than B and describes n units more than that of B acquiring the same speed, then

  1. a. (f + h)m2 = fhn
  2. b. (f – fh)m2 = fhn
  3. c. (h – f)n = fhm2
  4. d. (1/2)(f + h)n =fhm2

Solution:

  1. Answer: c

    S + n = (1/2) f(t + m2)and S =(1/2)ht2 , V = ht

    (1/2)ht2 +n= (1/2) f(t + m2) ……….(1)

    Also V = 0 + ht = 0 + f(t +m) ⇒ t + m =(ht/f) (put in equation (1))

    from equation (1)

    12ht2+n=12f(htf)2\Rightarrow \frac{1}{2} \mathrm{ht}^{2}+\mathrm{n}=\frac{1}{2} \mathrm{f}\left(\frac{\mathrm{ht}}{\mathrm{f}}\right)^{2}

    t2=2nfh(hf)\Rightarrow \mathrm{t}^{2}=\frac{2 \mathrm{nf}}{\mathrm{h}(\mathrm{h}-\mathrm{f})}

    Also,

    ht=f(t+m)t2=m2f2(hf)2\mathrm{ht}=\mathrm{f}(\mathrm{t}+\mathrm{m}) \Rightarrow \mathrm{t}^{2}=\frac{\mathrm{m}^{2} \mathrm{f}^{2}}{(\mathrm{h}-\mathrm{f})^{2}}

    Therefore, 2nfh(hf)=m2f2(hf)2\frac{2 \mathrm{nf}}{\mathrm{h}(\mathrm{h}-\mathrm{f})}=\frac{\mathrm{m}^{2} \mathrm{f}^{2}}{(\mathrm{h}-\mathrm{f})^{2}}

    n(hf)=12fhm2\Rightarrow \mathrm{n}(\mathrm{h}-\mathrm{f})=\mid \frac{1}{2} \mathrm{f} \mathrm{hm}^{2}


Question 70: The area bounded by y = x + 1 and y = cos x and the x-axis, is

  1. a. 1 sq. unit
  2. b. (3/2) sq. unit
  3. c. (1/4) sq. unit
  4. d. (1/8) sq. unit

Solution:

  1. Answer: b

    WBJEE 2019 Mathematics Practice Paper

    Area=12×1×1+0π/2cosxdxArea=\frac{1}{2}\times 1\times 1+\int_{0}^{\pi /2}cosxdx

    =12+sinx0π/2=\frac{1}{2}+\left.\sin x\right|_{0} ^{\pi / 2}

    = (1/2)+1= 3/2


Question 71: Let x1, x2 be the roots of x2 – 3x + a = 0 and x3, x4 be the roots of x2 – 12x + b = 0. If x1< x2< x3< x4 and x1, x2, x3, x4 are in G.P. then ab equals

  1. a. 24/5
  2. b. 64
  3. c. 16
  4. d. 8

Solution:

  1. Answer: b

    WBJEE 2019 Mathematics Practice Question Paper

    Since x1 , x2, x3, x4 are in G.P, then

    x3+x4x1+x2=123\frac{x_{3}+x_{4}}{x_{1}+x_{2}}=\frac{12}{3}

    Ar2+Ar3A+Ar=4\frac{Ar^{2}+Ar^{3}}{A+Ar}=4

    ⇒r2 = 4 ⇒ x = 2 (since G.P is increasing)

    ⇒x1 + x2 = 3

    ⇒A + Ar = 3

    ⇒ A(3) = 3

    ⇒A = 1

    Therefore, ab = x1 • x2 • x3 • x4 = 1.2. 4.8 = 64


Question 72: If q ϵ\epsilon R and 1icosθ1+2icosθ\frac{1-icos\theta }{1+2icos\theta } is real number, then θ will be (when I : Set of integers)

  1. a. (2n + 1) (π/2) , n ϵI\epsilon I
  2. b.3nπ/2 , n ϵ\epsilon I
  3. c. nπ, n ϵ\epsilon I
  4. d. 2nπ, n ϵ\epsilonI

Solution:

  1. Answer: a

    Using rationalization

    Practice Question Paper of WBJEE 2019 Mathematics


Question 73: Let A = (303030303)\begin{pmatrix} 3 & 0 &3 \\ 0& 3 &0 \\ 3& 0 & 3 \end{pmatrix}